Uniform Convergence: Showing Continuous & Uniformly Continuous

[mattmns]

Here is the question from the book:
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Let f: \mathbb{R} \to \mathbb{R} be a function. For any a\in \mathbb{R}, let f_a :\mathbb{R}\to \mathbb{R} be the shifted function f_a(x):=f(x-a).(a) Show that f is continuous if and only if, whenever (a_n)_{n=0}^{\infty} is a sequence of real numbers which converges to zero, the shifted functions f_{a_n} converge pointwise to f.

(b) Show that f is uniformly continuous if and only if, whenever (a_n)_{n=0}^{\infty} is a sequence of real numbers which converges to zero, the shifted functions f_{a_n} converge uniformly to f.---------
(\Rightarrow) Let x_0 \in \mathbb{R}. Suppose f is continuous. That is, given \epsilon > 0 there exists \delta > 0 such that if |x-x_0| < \delta then |f(x)-f(x_0)| < \epsilon.Let x_n = x_0 - a_n. So given \delta > 0 there exists N' > 0 such that |x_n - x_0|< \delta for all n > N'.

Given \epsilon' > 0 take \epsilon = \epsilon', and take N = N'.

So by the continuity of f we get that |x_n - x_0| < \delta for all n>N' = N.

Thus, |f(x_n) - f(x_0)| < \epsilon = \epsilon'.

But, f(x_n) = f(x_0-a_n) = f_{a_n}(x_0) for all n>N.

So we have |f_{a_n}(x_0) - f(x_0)| < \epsilon for all n>N.Thus f_{a_n} converges pointwise to f.
(\Leftarrow) Suppose given a_n \to 0 that f_{a_n} converges pointwise to f. Given x_0 take a sequence x_n which converges to x_0. That is, a_n = x_0 - x_n \to 0.

Since f_n converges pointwise to f, given \epsilon > 0 there is some N>0 such that |f_{a_n}(x_0) - f(x_0)| < \epsilon for all n>N.

But, |f_{a_n}(x_0) - f(x_0)| = |f(x_0 - a_n) - f(x_0)| = |f(x_n) - f(x_0)|.

Hence, |f(x_n) - f(x_0)| < \epsilon for all n>N.

That is, f(x_n) converges to f(x_0). Thus, f is continuous.----------

I don't think there is anything wrong with any of that (if there is, or you have any comments please say something).

However, I am not too sure about part (b).

The first direction is the same as in part (a), but the other direction I am not sure about.

I have been trying just the basic epsilon-delta definition way (similar to what I did in part (a) with this direction) and not really getting anywhere (which I guess could be expected since we don't have a uniform continuity definition with sequences [at least in our book]).

Any ideas? Thanks!edit... I am converting the $ signs to 's. Is there an easy (fast and painless) way to do this?

 

[StatusX]

Copy into notepad, search and replace, using pre-spaces to distinguish opening from closing ones. If it's not too late.

 

[StatusX]

What is your definition of uniform continuity? (Please always post relevant definitions)

 

[mattmns]

Cool, that worked nicely! Any ideas about the exercise now

Sorry, I thought that was a standard definition. Here it is:

Uniform Continuity: Let f:X\to Y be a map from one metric space (X,d_X) to another (Y,d_Y). Then f is uniformly continuous if, for every \epsilon > 0, there exists a \delta > 0 such that d_Y(f(x),f(x')) < \epsilon whenever x,x'\in X are such that d_X(x,x')<\delta.

Here is our book's definition of uniform convergence:

Let (f^{(n)})_{n=1}^{\infty} be a sequence of functions from one metric space (X,d_X) to another (Y,d_Y), and let f: X\to Y be another function. We say that (f^{(n)})_{n=1}^{\infty} converges uniformly to f on X if for every \epsilon > 0 there exists N>0 such that d_Y(f^{(n)}(x),f(x))<\epsilon for every n>N and x\in X.

 

[StatusX]

It is standard, it's just a lot easier to refer to a definition that's here. Remeber that the definition of continuity in terms of sequences was just a result of the more general definition of limits in terms of sequences, and you should still be able to apply that.